Let $f$ be a complex-valued function of a complex variable. Let $\Omega$ be the set of points where $f$ is holomorphic -- that is, the set of points $z\in\mathbb{C}$ where $f$ is holomorphic in an open neighborhood of $z$.
Take $z_0\in\Omega$, and let $\delta$ be the distance from $z_0$ to the boundary of $\Omega$. Let $z_s$ be one of the points on the boundary of $\Omega$ at distance $\delta$ from $z_0$, so that $|z_0-z_s|=\delta$. Suppose that the Taylor series for $f$ at $z_0$ has radius of convergence $\delta$, and consider the Taylor series expansions of $f$ centered at points on the line segment from $z_0$ to $z_s$, $\overline{z_0z_s}$.
Must the Taylor series of $f$ centered at $z_1\in\overline{z_0z_s}$ have radius of convergence exactly equal to $\delta-|z_1-z_0|$, or can it be greater?
Note: contrary to the familiar slogan that "the radius of convergence of the Taylor series at $z_0$ is the distance from $z_0$ to the closest singularity of $f$," there is a difference between the disk on which the Taylor series converges, and the disk on which the Taylor series converges to $f$. The second disk is contained in the first, but the two disks need not have the same radius. (Ahlfors makes this point clearly at the top of page 178: "The radius of convergence of the Taylor series is thus at least equal to shortest distance from $z_0$ to the boundary of $\Omega$. It may well be larger.") In the setting of the question, it is obvious that the Taylor series at a point $z_1\in\overline{z_0z_s}$ can't converge to $f$ past $z_s$. But that alone does not mean the series at $z_1$ can't converge, period, past $z_s$.
What I'm asking is whether the fact that the Taylor series at $z_0$ doesn't converge past $z_s$ implies the same for all other Taylor series centered at points on the radius connecting $z_0$ and $z_s$, or whether, by contrast, we might get a Taylor series that converges past $z_s$ if we center the Taylor series at a point closer to $z_s$.
The question does not seem trivial, but perhaps I am missing something. Rudin, at least, points out in PMA (page 177) that when we change the center of a power series with disk of convergence $|x|<R$ to $x=a$, the result "may actually converge in a larger interval than $|x-a|<R-|a|$."
Define $f(z) = 1/(1-z), z\in \mathbb D, f(z) = |z|, |z|\ge 1.$ Then $\mathbb D$ is our $\Omega $ here. Let $z_0 = 0, z_s = -1.$ Then the radius of convergence of the power series of $f$ at $z_0$ is $1.$ But the radius of convergence of the power series of $f$ at $-1/2$ is $3/2.$
Added later in response to your comment: Yes, you are right that there will be such a point on the boundary. (And as $1/(1-z)$ shows, it's possible that there is only one such point.)
For simplicity, I'll work in $\mathbb D.$ Suppose $f(z) = \sum a_nz^n$ in $\mathbb D,$ where the power series has radius of convergence (ROC) $1.$
Claim: There exists $\zeta \in \partial \mathbb D$ such that for all $a\in [0,\zeta),$ the ROC of the Taylor series of $f$ at $a$ equals $1-|a|.$
Proof: If not, then for all $\zeta \in \partial \mathbb D,$ there exists $a_{\zeta} \in [0,\zeta)$ where the ROC, lets call it $r_{\zeta},$ is greater than $1-|a_{\zeta}|.$ It follows that for every $\zeta,$ $f$ extends to a holomorphic $f_\zeta$ on $\mathbb D \cup B(a_\zeta, r_{\zeta} ).$
OK, I'll be brief from here. Using compactness of $\overline {\mathbb D},$ and the identity principle (which tells you the $f_\zeta$'s agree on any overlap), you can see that $f$ extends analytically to a holomorphic function on $D(0,R)$ for some $R>1.$ The Taylor series at $0$ for the extension has ROC $\ge R,$ but this is exactly the original power series for which the ROC = 1. This contradiction gives your result.